Using [1]}, the sum \(A+B\)  is maximally monotone and hence, its graph is closed in \(\mathcal {H}^{weak}\times \mathcal {H}^{strong}\)  [1]}.
Therefore, \(0\in (A+B) x^*\)  a.s., that is \(x^*\in \mathcal {P}\)  a.s. By Lemma REF , the sequence \((x_n)_{n \in \mathbb {N}}\)  converges weakly to \(\bar{x} \in \mathcal {P}\)  a.s. and the proof is completed.
